Number System Practice Gym by Anubhav Sehgal - Part 1


  • NMIMS, Mumbai (Marketing)


    Q1) Two numbers when divided by m(m < = 7) gives remainder 3 and 4. The remainder when the sum of these two numbers is divided by the same divisor is R. Find the value of R + m.

    Let the two numbers be a and b
    a = mq + r1
    b = nq + r2
    Then, (a + b) = (m + n)q + (r1 + r2)
    Remainder when sum of two numbers (giving individual remainders r1 and r2 with a divisor) is divided by the same divisor is :
    R = (r1 + r2) - divisor [if this is positive]
    else
    R = (r1 + r2)
    Here R = (3 + 4) - m
    R + m = 7

    Q2) Calculate GCD(1,155) + GCD(2,155) + .. + GCD(155,155). where GCD(m,n) is the greatest common divisor of m and n.

    GCD(x,155) = 1 unless x is a multiple of 5 and/or 31
    4 only-multiples of 31
    30 only-multiples of 5
    1 multiple of both 31 and 5
    35 such numbers
    Hence,
    Sum = 4 * 31 + 30 * 5 + 1 * 155 + (155 - 35) * 1 = 549

    This is an example of how test setters combine two simple concepts to trap you.

    Q3) Find the largest number n such that (n - 11) divides n^3 + 83.

    A mix of elementary algebra and number system
    By remainder theorem,
    Remainder when n^3 + 83 is divided by (n - 11) = 11^3 + 83 = 1414
    Now,
    (n - 11) * k = 1414
    For largest n,
    k = 1, (n - 11) = 1414
    n = 1425

    Q4) Consider natural numbers a, b, c, d and e.Three statements are provided:
    I. (de)^2 = abbb
    II.(ee)^2 = ccbb
    III. A perfect square never ends with c,d and e.

    Which of the following statements are sufficient to determine the arithmetic mean of a,b and c?
    a) Any of the three statements
    b) Any two of the three statements
    c) Statement I and II are sufficient
    d) Statement II and III are sufficient
    e) All three statements are necessary

    Square with largest occurrence of a non-zero digit at the end is 38^2 = 1444
    The only square of the form XX^2 = YYZZ is 88^2 = 7744
    A square never ends with 2,3,7,8 or odd number of zeroes
    Combine them, analyze the options and you ll find I and II are sufficient to answer what we need which is primarily values of a,b,c.

    Q5) Find 24^1202 mod 1446.

    0_1490618109343_q5_sol.jpg

    Q6) Find the smallest whole number that when divided by 5,7,9 and 11 leaves a remainder of 1,2,3 and 4 respectively.

    If the difference between divisors and respective remainders was equal you simply apply : LCM(divisors) - common difference. But here you need to tune the formula and that can be done when you know how the standard formula comes.

    If N was your original number where
    N = 5a + 1 = 7b + 2 = 9c + 3 = 11d + 4
    then :
    0_1490618324258_q6_sol.jpg

    Q7) Find the highest N such that 11^N divides (97! + 98! + 99!)

    OA : 10, Guess doesn't need discussion :)

    Q8. When a number N(N > 500) is successively divided by 3,5,7, it leaves a remainder of 2,3,6 respectively. Find the remainder when 2nd such smallest N is divided by 35.

    N = 3(5(7m + 6) + 3) + 2 = 105m + 101
    N = 101,206,311,416,521,626
    626 mod 35 = 31

    Q9. Find the highest power of 1001 in 1001 * 999 * 997 * 995 * ... * 3 * 1

    Approach 1 :
    Multiply and divide by 2 * 4 * 6 * 8 *... * 998 * 1000
    Number = 1001!/(2 * 4 * 6 * 8 * .. * 998 * 1000) = 1001!/2^500 * 500!
    Exponent of 13 in this = (77 + 5) - (38 + 2) = 42

    Approach 2 :
    For power of 13, 13 * 1 to 13 * (7 * 11) all odd powers
    So, (7 * 11 + 1)/2 = 39
    For 13^2,
    13 * (13 * 1), 13 * (13 * 3), ... up-to 13 * (13 * 5)
    So additional three powers.
    Total = 39 + 3 = 42

    Q10) If p, q, and r are positive integers such that (p + q - r)(q + r - p)(p + r - q) = 15, then what is the product of p, q and r?
    a) 24
    b) 60
    c) 64
    d) Cannot Be Determined

    OA : CBD
    15 can be broken into three factors as follows - (5, 3, 1) or (15, 1, 1).
    Solving for p, q, r we get (4, 3, 2) and (8, 8, 1) as the triplets corresponding to the above set of factors.
    Thus product of p, q and r can be either 24 or 64.

    Q11) Sum of two numbers and their LCM is p where p is a prime and 10 < p < 30. For how many p do we have 3 pairs of such number?

    Let the numbers be ha and hb.
    LCM = hab since LCM * HCF = ha * hb
    So,
    h(a + b + ab) = p
    Case 1 : h != 1
    Then, h = p => prime => Not possible
    Since LCM >= Numbers >= HCF always
    Case 2 : h = 1
    a + b + ab = p
    (a + 1)(b + 1) = (p + 1)
    Number of pairs = [Factors of (p + 1)]/2 - 1 for the case when a becomes zero.
    For the given range,
    p = 11, 13, 17, 19, 23, 29
    p = 11 => 12 = 2^2 * 3 => 3 * 2/2 - 1 = 2 pairs
    p = 13 => 14 = 2 * 7 => 2 * 2/2 - 1 = 1 pair
    p = 17 => 18 = 2 * 3^2 => 2 pairs
    p = 19 => 20 = 2^2 * 5 => 2 pairs
    p = 23 => 24 = 2^3 * 3 => 4 * 2/2 - 1 = 3 pairs
    p = 29 => 30 = 2 * 3 * 5 => 2 * 2 * 2/2 - 1 = 3 pairs
    So 2 such p = 23,29

    Q12) A natural number N is having a total of 21 composite factors. Let X be the maximum number of prime factors of N then find the remainder when X is divided by 3.

    X = 2. Hence remainder X mod 3 = 2

    0_1490618744197_q12_sol.jpg


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